To prove the question what I did: for an unknown random variable $U$ $$E(Y\mid E(X))=E(E(U\mid X)\mid E(X))=E(E(U\mid E(X))\mid X),$$ because $E(X)$ is function of X. So, finally $$E(Y\mid E(X))=E(U\mid E(X))=E(Y).$$ The problems are here: I have to assume $Y=E(U\mid X)$ and $Y=U\mid E(X),$ which may be possible if there is a linear function such that $U=f(X).$ Is my idea right? If not then do you have any ideas or recommendations? And how do I find the variable $U$ so that it has a well defined relation with $X$ and $Y$.
The above question arises from a regression model can be described in a different way. Suppose, I have a formula based on E(X) but I can only use E(Y). However, I know there is a linear relationship between Y and X such that I can formulate a regression model. For example for a regression model we have $$Y=\alpha + \beta X + \text{error},$$ therefore, $$E(Y\mid X)=\alpha+\beta X.$$ So, We can have $$E(Y\mid E(X))=\alpha+\beta E(X),$$ for a value $E(X).$ My question is, since I can not use $E(X)$, then what is the corresponding value of $E(Y)$ should I use instead of $E(X)$?
$E(X)$ is just some number, or constant function if you like. The $\sigma$-field $\mathcal{F}$ generated by this constant function is trivial. Hence, $E(Y \mid E(X)) = E(Y \mid \mathcal{F}) = E(Y)$.
See this question: Conditional Expectation: What happens if you take conditional expectation on trivial sigma field?